8 coins are simultaneously flipped.  What is the probability that heads are showing on at most 2 of them?
Solution: The number of ways for the outcome to have exactly 0, 1, or 2 heads are $\binom{8}{0}=1$, $\binom{8}{1}=8$, or $\binom{8}{2}=28$, respectively.  There are $2^8$ total possible outcomes (2 possibilities for each coin, and 8 coins). So the answer is $\dfrac{1+8+28}{2^8}=\boxed{\dfrac{37}{256}}$.